Show that
cot((x+y)/2)=-(sin(x)-sin(y))/(cos(x)-cos(y))
for all values of x and y for which both sides are defined.
using the half-angle formulas,
sin(x)-sin(y) = 2 cos((x+y)/2)sin((x-y)/2)
cos(x)-cos(y) = -2sin((x+y)/2)sin((x-y)/2)
ta-da!
Wow, I tried that but I messed up signs! Thanks!
Oops I meant sum and difference
No, I meant sum and difference. I started out trying the half-angle for tangent, but got bogged down. Then I saw the sums on the right and went the easy way.
To prove that cot((x+y)/2)=-(sin(x)-sin(y))/(cos(x)-cos(y)), we need to simplify and manipulate the expression on the right side until it matches the left side.
1. Start by using the double-angle identity for cotangent:
cot((x+y)/2) = (cos((x+y)/2))/(sin((x+y)/2))
2. Next, we will simplify the expression on the right side by using the sum-to-product identities:
sin(x) - sin(y) = 2sin((x-y)/2)cos((x+y)/2)
cos(x) - cos(y) = -2sin((x+y)/2)sin((x-y)/2)
Plugging these values back into the expression gives:
cot((x+y)/2) = (cos((x+y)/2))/(sin((x+y)/2))
= 2sin((x-y)/2)cos((x+y)/2)/(-2sin((x+y)/2)sin((x-y)/2))
3. Cancel out the common factors of 2 and -2:
cot((x+y)/2) = -sin((x-y)/2)cos((x+y)/2)/(sin((x+y)/2)sin((x-y)/2))
4. Now, we can simplify the expression further by canceling out the common factors of sin((x-y)/2):
cot((x+y)/2) = -cos((x+y)/2)/sin((x+y)/2)
5. Finally, using the reciprocal identity for cotangent:
cot((x+y)/2) = -(sin((x+y)/2))/(cos((x+y)/2))
Notice that we have arrived at the left side of the equation, which proves that cot((x+y)/2)=-(sin(x)-sin(y))/(cos(x)-cos(y)).
Therefore, we have shown that cot((x+y)/2)=-(sin(x)-sin(y))/(cos(x)-cos(y)) for all values of x and y for which both sides are defined.